1. Multivariable Optimization
ECON 1150
Spring 2013
Lecture 5
Multivariable Optimization
In the previous chapter, all functions involved depend on one variable only. In reality, there are many variables in a function. Since it is too complicated to study all factors at the same time, we use the ‘other things being equal’ strategy.
ECON 1150, Spring 2013

2. 1. Necessary Conditions Optimization problems: maxx1,x2 y = f(x1,x2)
ECON 1150
Spring 2013
1. Necessary Conditions
Optimization problems:
maxx1,x2 y = f(x1,x2)
minx1,x2 y = f(x1,x2)
E.g., profit maximization, cost minimization
ECON 1150, Spring 2013

3. f(x1, c) is a function of x1 only. When x1 is kept constant,
ECON 1150
Spring 2013
When x2 is kept constant,
f(x1, c) is a function of x1 only.
When x1 is kept constant,
f(c, x2) is a function of x2 only
First-order conditions for a stationary point
In the case of functions of one variable, not all stationary points need give extreme values. In the case of functions of n variables, there is the further possibility that a stationary point may be a saddle point.
Show the graph with a horizontal plane.
ECON 1150, Spring 2013

4. Example 5.1: Find stationary values of the following functions:
ECON 1150
Spring 2013
Example 5.1: Find stationary values of the following functions:
(a) y = 2x1² + x2²;
(b) y = 4x1² - x1x2 + x2² - x1³.
y1 = 4x1 = 0 => x1 = 0; y2 = 2x2 = 0 => x2 = 0. y = 0
b. y1 = 8x1 – x2 – 3x12 = 0; y2 = -x1 + 2x2 = 0. x2 = 0.5x1.
8x1 – 0.5x1 – 3x12 = x1(7.5 – 3x1) = 0 => x1 = 0 or x2 = 0 or 1.25.
For (0, 0), y = 0.
For (2.5, 1.25), y = 4(2.52) – (2.5)(1.25) – 2.53 =
ECON 1150, Spring 2013

5. Possible cases of stationary points
ECON 1150, Spring 2013

6. 2. Sufficient Conditions
ECON 1150
Spring 2013
2. Sufficient Conditions
Second-order conditions for a local maximum at a point (SOC max)
f11 < 0; f11·f22 – (f12)2 > 0
Second-order conditions for a local minimum at a point (SOC min)
f11 > 0; f11·f22 – (f12)2 > 0.
f11(x1*,x2*)f22(x1*,x2*) – (f12(x1*,x2*))2 > 0
SOC max  f22 < 0
SOC min  f11 > 0
Remarks for SOC:
They refer to a certain point only
They are sufficient, but not necessary
They assume FOC are satisfied.
ECON 1150, Spring 2013

7. Classifying a stationary point
ECON 1150
Spring 2013
Classifying a stationary point
If f1(x1*,x2*) = f2(x1*,x2*) = 0, then
a. SOC max  local maximum
SOC min  local minimum
f11f22 – (f12)2 = 0  no conclusion
f11f22 – (f12)2 < 0  saddle point
ECON 1150, Spring 2013

8. ECON 1150
Spring 2013
Example 5.2: Identify the nature of the stationary points of the following function,
y = f(x1,x2) = x1³ + 5x1x2 – x2².
Theorem states necessary conditions only, not sufficient conditions.
y1 = 3x12 + 5x2 = 0
y2 = 5x1 – 2x2 = 0
x2 = 2.5x1
(3x )x1 = 0 => x1 = 0 or x1 = -25/6.
x2 = 0 or -125/12.
f11 = 6x1, f22 = -2, f12 = f21 = 5
 = f11* f22 – (f12)2 = -12x1 – 25
For x1 = x2 = 0,
f11 = 0, f22 = -2
 = => saddle point
For x1 = -25/6 and x2 = -125/12,
f11 = 6(-25/6) = -25 < 0, f22 = -2 < 0
 =-12(-25/6) – 25 = 50 – 25 = 25 > 0 => local maximum
ECON 1150, Spring 2013

9. ECON 1150
Spring 2013
Example 5.3: Identify the nature of the stationary points of the following functions:
f(x1,x2) = x12 + x24;
f(x1,x2) = x12 – x24;
f(x1,x2) = x12 + x23;
f(x1,x2) = x14 + x24;
f1 = 2x1, f2 = 4x23,
FOC: f1 = f2 = 0 => x1 = x2 = 0
f11 = 2, f22 = 12x22, f12 = f21 = 0
f11 x f22 – (f12)2 = 0
ECON 1150, Spring 2013

10. then f(x1,x2) is a concave function.
If for all (x1,x2),
f11  0 and f11f22 – (f12)2  0.
then f(x1,x2) is a concave function.
Example 5.4: Concave functions
y = x1 + x2;
y = x10.4x for positive x1 and x2;
y = lnx1 + lnx2.
ECON 1150, Spring 2013

11. then f(x1,x2) is a convex function.
If for all (x1,x2),
f11  0 and f11f22 – (f12)2  0.
then f(x1,x2) is a convex function.
Example 5.5: Convex functions
y = x1 + x2;
y = 3x12 + 3x1x2 + x22.
ECON 1150, Spring 2013

12. For a stationary point (x10, x20), concave function  global maximum
ECON 1150
Spring 2013
For a stationary point (x10, x20),
concave function  global maximum
convex function  global minimum
Example 5.6: Show that the function
f(x1,x2)
= -2x12 – 2x1x2 – 2x x1 + 42x2 – 158
is a concave function. Find the global maximum point.
f1 = -4x1 – 2x = 0
f2 = -2x1 – 4x = 0
=> (x1,x2) = (5,8).
f11 = -4, f12 = f21 = -2, f22 = -4
f11f22 – (f12)2 = (-4)(-4) – (-2)2 = 12 > 0
Therefore it is a concave function
ECON 1150, Spring 2013

13. a. 2nd order conditions hold at a point  Local extremum
Remarks:
a. 2nd order conditions hold at a point
 Local extremum
b. 2nd order conditions hold for all points
 Global extremum
c. SOC are sufficient, but not necessary
ECON 1150, Spring 2013

14. 3. Economic Applications
Spring 2013
3. Economic Applications
A firm produces 2 different kinds A and B of a commodity. The daily cost of producing x units of A and y units of B is
C(x,y) = 0.04x xy y2 + 4x + 2y
Suppose that that firm sells all its output at a price per unit of 15 for A and 9 for B. Find the daily production levels x* and y* that maximize profit per day.
Textbook: Example 2 on p. 465.
Profit per day is (x,y) = 15x + 9y – C(x,y). So
(x,y) = 15x + 9y – 0.04x2 – 0.01xy – 0.01y2 – 4x – 2y – 500
= x2 – 0.01xy – 0.01y2 + 11x + 7y – 500.
The first-order partial derivatives are
x(x,y) = -0.08x – 0.01y and y(x,y) = -0.01x – 0.02y + 7.
If x* > 0 and y* > 0 maximize profit, then by the first-order condition, x = y = 0 at (x*, y*). Hence,
-0.08x* y* + 11 = (1)
-0.01x* y* = (2)
(2) – 2(1),
0.15x* - 15 = 0  x* = 100.
Put x* = 100 into (2),
-1 – 0.02y* + 7 = 0
0.02y* = 6  y* = 300.
The second-order partial derivatives are
xx = -0.08, yy = -0.02, xy = yx = Then
 = xxyy – (xy)2 = (-0.08)(-0.02) – (-0.01)2 = > 0.
Since xx < 0 and  > 0 for all x and y, the second-order conditions for a global maximum are satisfied.
ECON 1150, Spring 2013

15. Long-run profit maximization of a competitive firm Output price: 200
ECON 1150
Spring 2013
Long-run profit maximization of a competitive firm
Output price: 200
Inputs: K with a price of 42
L with a price of 5
Production function: 3.1K0.3L0.25.
Profit:  = 200(3.1K0.3L0.25) – 42K – 5L = 620K0.3L0.25 – 42K – 5L.
FOC:
K = 186K-0.7L0.25 – 42 = (1)
L = 155 K0.3L-0.75 – 5 = (2)
(1) => L0.25 = (42/186) K0.7
Raise to the power 3,
L0.75 = (423/1863)K (3)
(2) => L0.75 = 31K (4)
Combine (3) and (4),
(423/1863)K2.1 = 31K  K1.8 = 31 * 1863 /  K =
(1) / (2) =>
(186/155) * L / K = 42 / 5  L = 42 * 31 * K / 186 =
KK = 130.2K-1.7L0.25 = < 0
KL = 46.5K-0.7L-0.75 =
LL = K0.3L-1.75 =
KK  LL – (KL)2 = ( )( ) – ( )2 =
The second order conditions for a local maximum are satisfied.
ECON 1150, Spring 2013

16. General profit maximization problem
ECON 1150
Spring 2013
General profit maximization problem
maxK,L (K,L) = pf(K,L) – wKK – wLL
FOC:
K(K,L) = pMPK – wK = 0
L(K,L) = pMPL – wL = 0
pMPn = wn (n = K,L)
wK / MPK = wL / MPL = p = MC
wK/wL = MPK/MPL = MRTS
pMP is called the value of marginal product.
ECON 1150, Spring 2013

17. Profit maximization of a 2-product firm
ECON 1150
Spring 2013
Profit maximization of a 2-product firm
A two-product firm faces the demand and cost functions below:
Q₁ = P₁ - P₂
Q₂ = 35 - P₁ - P₂
TC = Q₁² + 2Q₂² + 10
Find the profit-maximizing output levels and the maximum profit.
Q1 – Q2 = 5 – P1 => P1 = 5 – Q1 + Q2
Q1 – 2Q2 = P2 => P2 = 30 + Q1 – 2Q2
TR1 = P1Q1 = (5 – Q1 + Q2)Q1 = 5Q1 – Q12 + Q1Q2
TR2 = P2Q2 = (30 + Q1 – 2Q2)Q2 = 30Q2 + Q1Q2 – 2Q22
Profit:  = TR1 + TR2 – TC = (5Q1 – Q12 + Q1Q2) + (30Q2 + Q1Q2 – 2Q22) – (Q12 + 2Q )
= -2Q12 – 4Q22 + 2Q1Q2 + 5Q1 + 30Q2 – 10
FOC:
1 = -4Q1 + 2Q2 + 5 = 0
2 = -8Q2 + 2Q = 0
Rearranging terms,
4Q1 – 2Q2 = (1)
2Q1 – 8Q2 = (2)
4(1) – (2),
14Q1 = => Q1* = 25 / 7
(1) – 2(2),
14Q2 = 65 => Q2* = 65 / 14
SOC:
11 = -4, 22 = -8, 12 = 21 = 2
1122 – (12)2 = (-4)(-8) – 22 = 28 > 0 (max)
Since 11 < 0 and 1122 – (12)2 > 0 for all Q1 and Q2, the profit is a concave function. Thus Q1* and Q2* are the global profit-maximizing output levels.
* = -2(25/7)2 – 4(65/14)2 + 2(25/7)(65/14) + 5(25/7) + 30(65/14) – 10 =
ECON 1150, Spring 2013